I'm looking for a little help on this one. I've been looking at it for hours. An airplane whose air speed is 620km/h is supposed to fly in a straight path 35.0 degrees north to east. But a steady 95 km/h wind is blowing from the north. In what direction should the plane head? I know that 95 km/h is the y value. I suspected that 620 was the resultant (not sure on this) and that the angle was 35 degrees. Looking at it I know that the angle must be greater than 35 degrees since there is a force of 95 km/h bearing down on the plane. My thought : use a^2 + b^2 = c^2 solve for b and calculate the angle using tan-1(95/613)= 8.8 Using common sense I know that 8.8 degrees isn't right. The answer is 42.2 North of East. Any help is appreciated.